Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 7;
const int LG = 20;
const int LEN = 10;
const int INF = 1e9 + 7;
vector<int> a[MAXN];
int n, m, q;
vector<int> g[MAXN];
void read() {
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; ++i) {
    int u;
    cin >> u;
    a[u - 1].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    sort(a[i].begin(), a[i].end());
    if (a[i].size() > LEN) a[i].resize(LEN, INF);
  }
}
int tin[MAXN], tout[MAXN], timer = 0;
int go[MAXN][LG];
vector<int> dp[MAXN][LG];
void dfs(int u, int parent) {
  tin[u] = timer++;
  go[u][0] = parent;
  for (int i = 1; i < LG; ++i) go[u][i] = go[go[u][i - 1]][i - 1];
  if (u == 0) {
    for (int i = 0; i < LG; ++i) {
      dp[u][i] = a[u];
    }
  } else {
    dp[u][0] = a[u];
    for (int i = 1; i < LG; ++i) {
      vector<int> t(dp[u][i - 1].size() + dp[go[u][i - 1]][i - 1].size());
      merge(dp[u][i - 1].begin(), dp[u][i - 1].end(),
            dp[go[u][i - 1]][i - 1].begin(), dp[go[u][i - 1]][i - 1].end(),
            t.begin());
      sort(t.begin(), t.end());
      if (t.size() > LEN) t.resize(LEN, INF);
      dp[u][i] = t;
    }
  }
  for (int v : g[u]) {
    if (v == parent) continue;
    dfs(v, u);
  }
  tout[u] = timer++;
}
bool anc(int u, int v) { return tin[u] <= tin[v] && tout[v] <= tout[u]; }
int lca(int u, int v) {
  if (anc(u, v)) return u;
  for (int i = LG - 1; i >= 0; --i) {
    if (!anc(go[u][i], v)) u = go[u][i];
  }
  return go[u][0];
}
void add(int u, vector<int> &ans, int k) {
  vector<int> t(ans.size() + a[u].size(), INF);
  merge(ans.begin(), ans.end(), a[u].begin(), a[u].end(), t.begin());
  if ((int)t.size() > k) t.resize(k, INF);
  ans = t;
}
void add(int u, int v, vector<int> &ans, int k) {
  if (anc(u, v)) return;
  for (int i = LG - 1; i >= 0; --i) {
    if (!anc(go[u][i], v)) {
      vector<int> t(ans.size() + dp[u][i].size());
      merge(ans.begin(), ans.end(), dp[u][i].begin(), dp[u][i].end(),
            t.begin());
      ans = t;
      u = go[u][i];
    }
  }
  add(u, ans, k);
}
void solve() { dfs(0, 0); }
vector<int> get(int u, int v, int k) {
  vector<int> ans;
  add(u, v, ans, k);
  add(v, u, ans, k);
  int l = lca(u, v);
  add(l, ans, k);
  return ans;
}
void print() {
  for (int i = 0; i < q; ++i) {
    int u, v, k;
    cin >> u >> v >> k;
    --u;
    --v;
    vector<int> ans = get(u, v, k);
    vector<int> t;
    for (int i = 0; i < (int)ans.size() && ans[i] < INF; ++i)
      t.push_back(ans[i]);
    cout << t.size() << ' ';
    for (int e : t) cout << e + 1 << ' ';
    cout << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
  print();
  return 0;
}